Maximum Units on a Truck

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ArraysGreedy Algorithms

You are assigned to put some amount of boxes onto one truck. You are given a 2D array boxTypes, where boxTypes[i] = [numberOfBoxes_i, numberOfUnitsPerBox_i]:

numberOfBoxes_i is the number of boxes of type i.
numberOfUnitsPerBox_i is the number of units in each box of the type i.

You are also given an integer truckSize, which is the maximum number of boxes that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed truckSize.

Return the maximum total number of units that can be put on the truck.

Example 1:

Input: boxTypes = [[1,3],[2,2],[3,1]], truckSize = 4
Output: 8
Explanation: There are:
- 1 box of the first type that contains 3 units.
- 2 boxes of the second type that contain 2 units each.
- 3 boxes of the third type that contain 1 unit each.
You can take all the boxes of the first and second types, and one box of the third type.
The total number of units will be = (1 * 3) + (2 * 2) + (1 * 1) = 8.

Example 2:

Input: boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10
Output: 91

Maximum Units on a Truck

Easy

Google

1 view

Topics:

ArraysGreedy Algorithms

You are assigned to put some amount of boxes onto one truck. You are given a 2D array boxTypes, where boxTypes[i] = [numberOfBoxes_i, numberOfUnitsPerBox_i]:

numberOfBoxes_i is the number of boxes of type i.
numberOfUnitsPerBox_i is the number of units in each box of the type i.

You are also given an integer truckSize, which is the maximum number of boxes that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed truckSize.

Return the maximum total number of units that can be put on the truck.

Example 1:

Input: boxTypes = [[1,3],[2,2],[3,1]], truckSize = 4
Output: 8
Explanation: There are:
- 1 box of the first type that contains 3 units.
- 2 boxes of the second type that contain 2 units each.
- 3 boxes of the third type that contain 1 unit each.
You can take all the boxes of the first and second types, and one box of the third type.
The total number of units will be = (1 * 3) + (2 * 2) + (1 * 1) = 8.

Example 2:

Input: boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10
Output: 91

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the number of box types and the truck size?
Can the number of boxes per box type or the number of units per box type be zero?
If the truck cannot hold any boxes, should I return 0?
If there are multiple box types that would maximize the units loaded, is the order in which I consider them significant, or can I choose any such combination?
Are the inputs always valid (e.g., positive integers), or should I handle potential edge cases like null or negative inputs?

Brute Force Solution

Approach

The brute force approach for this problem involves exploring every single possible way to load boxes onto the truck. We try every combination of box types until we find the best one that fits within the truck's capacity.

Here's how the algorithm would work step-by-step:

Consider each type of box one at a time.
Try loading zero boxes of the first type, then one box, then two boxes, and so on, up to the maximum number of boxes of that type available.
For each quantity of the first type of box you try, repeat the process for the second type of box: zero boxes, one box, two boxes, and so on.
Continue this pattern for all box types, trying all possible quantities of each.
For each combination of box quantities, calculate the total number of units you can load onto the truck, and check if the total number of boxes loaded fits within the truck's size limit.
Keep track of the combination that yields the most units while still fitting within the truck's capacity.
After considering all possible combinations, select the combination with the highest number of units as the answer.

Code Implementation

def maximum_units_brute_force(box_types, truck_size):
    maximum_units = 0

    def solve(current_box_index, current_truck_size, current_units):
        nonlocal maximum_units

        # If the truck is full, update the max units and stop
        if current_truck_size == 0:
            maximum_units = max(maximum_units, current_units)
            return

        # If we have considered all box types, update and stop
        if current_box_index == len(box_types):
            maximum_units = max(maximum_units, current_units)
            return

        number_of_boxes, units_per_box = box_types[current_box_index]

        # Iterate through all possible number of boxes of this type
        for box_count in range(min(number_of_boxes, current_truck_size) + 1):

            # Recursively call to include or exclude the box type
            solve(
                current_box_index + 1,
                current_truck_size - box_count,
                current_units + box_count * units_per_box
            )

    solve(0, truck_size, 0)
    return maximum_units

Big(O) Analysis

Time Complexity

O(product of box quantities) – The provided approach iterates through all possible combinations of box quantities for each box type. If we let 'k' be the number of box types and 'q_i' be the quantity of boxes available for the i-th box type, then the algorithm explores up to q_1 * q_2 * ... * q_k combinations. Therefore, the time complexity is proportional to the product of the maximum quantities of each box type. This makes the algorithm extremely inefficient as the quantities increase.

Space Complexity

O(1) – The described brute force approach doesn't explicitly create any auxiliary data structures like lists, hash maps, or significant variables that scale with the input size. While the algorithm iterates through combinations of boxes, it only needs to maintain a few variables for counters and calculating the current units and boxes loaded. These variables consume a constant amount of memory regardless of the number of box types or the truck size, making the auxiliary space complexity constant. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

To maximize the units on the truck, we should prioritize loading boxes with the highest number of units per box. This approach avoids inefficiently filling the truck with boxes that hold fewer units and ensures we grab the most valuable boxes first. The goal is to greedily pick from the most unit-dense boxes until the truck is full.

Here's how the algorithm would work step-by-step:

First, sort the different box types based on the number of units each box contains, with the highest units per box at the top.
Next, start loading the truck with boxes from the top of the sorted list. This means grabbing as many boxes as you can of the type that gives you the most units each.
Keep adding boxes of that type until either you've run out of that type of box or the truck is full.
If the truck isn't full yet, move to the next box type on your sorted list (the one with the next-highest units per box) and repeat the process.
Continue grabbing as many boxes as you can of each type, in order from most units per box to least, until the truck is completely full.
Finally, calculate the total number of units on the truck by adding up the units from all the boxes you loaded.

Code Implementation

def maximum_units(box_types, truck_size):
    # Sort box types by units per box in descending order
    box_types.sort(key=lambda box: box[1], reverse=True)

    total_units = 0
    space_remaining = truck_size

    for box_count, units_per_box in box_types:
        # Load as many boxes of the current type as possible
        if space_remaining >= box_count:
            total_units += box_count * units_per_box
            space_remaining -= box_count

        # Truck is full, exit the loop
        else:
            total_units += space_remaining * units_per_box
            space_remaining = 0
            break

    return total_units

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this algorithm is the sorting of the boxTypes array based on the number of units per box. The sorting step uses a comparison-based sorting algorithm which typically has a time complexity of O(n log n), where n is the number of box types. After sorting, the algorithm iterates through the sorted array to load the truck, but this linear iteration (O(n)) is less costly than the sorting operation. Therefore, the overall time complexity is determined by the sorting step.

Space Complexity

O(1) – The dominant operation affecting space complexity is the sorting step. While the plain English explanation doesn't specify a particular sorting algorithm, many efficient sorting algorithms like heapsort or quicksort (in-place versions) can be implemented with O(1) auxiliary space. We are not creating any additional data structures that scale with the number of box types. Therefore, the auxiliary space used is constant regardless of the number of box types N.

Edge Cases

Case	How to Handle
Empty boxTypes array	Return 0 since there are no boxes to load.
truckSize is 0	Return 0 as the truck cannot carry any boxes.
All box types have 0 boxes available	Return 0 as there are no boxes to load, even if truckSize is non-zero.
A box type has a negative number of boxes	Treat the number of boxes as 0 for that type, or throw an exception, depending on problem constraints.
A box type has negative units per box	Treat the units per box as 0 for that type, or throw an exception, depending on problem constraints.
truckSize is larger than the total number of boxes available	Load all available boxes and return the total units.
Integer overflow when calculating total units	Use a data type with a larger range (e.g., long) to store the result.
Large number of box types, each with a small number of boxes and small units per box	Sorting dominates the runtime, but given the problem constraints, this may be acceptable. Consider specialized sorting if needed.